$$
\begin{aligned}
& \text { Normal vector }=\overrightarrow{\mathrm{AB}}=(\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}) \\
& =(6 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})
\end{aligned}
$$
Or $2(3 \hat{i}-2 \hat{j}+\hat{k}) \mid$
$$
\begin{aligned}
& P \equiv 3(x+1)-2(y)+1(z-2)=0 \\
& P \equiv 3 x-2 y+z+1=0 \\
& P^{\prime} \equiv 2 x+y+3 z-1=0
\end{aligned}
$$
angle between $P$ \& $P^{\prime}=\frac{\hat{n}_1 \cdot \hat{n}_2}{\left|n_1 \| n_2\right|}=\cos \theta$
$$
\begin{aligned}
& \theta=\cos ^{-1}\left(\frac{6-2+3}{\sqrt{14} \times \sqrt{14}}\right) \\
& \theta=\cos ^{-1}\left(\frac{7}{14}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}
\end{aligned}
$$
Option C is correct.